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On 15 April 2017 at 06:04, Alex Lee <[log in to unmask]> wrote:

Thanks Eleanor,
If I understand right, there is just 1 TFZ. TFZ== means "Translation Function Z-score equivalent, only calculated for the top solution after refinement (or for the number of top files specified by TOPFILES)" so there could be many TFZ==.

Twinning analysis attached below:
TWINNING ANALYSIS:
Global twinning statistics.
These tests rely on the fact that it is highly improbably that very weak or very strong reflections will coincide, therefore, the tails for the distribution of twinned datasets will be less pronounced
Data truncated to 67.93 - 3.50 A resolution
$TABLE: Cumulative intensity distribution:
$GRAPHS: Cumulative intensity distribution (Acentric and centric):N:1,2,3,4,5,6:
$$ Z Acent_theor Acent_twin Acent_obser Cent_theor Cent_obser $$
$$
 0.00000 0.00000 0.00000 0.01988 0.00000 -
 0.04000 0.03921 0.00303 0.03225 0.15852 -
 0.08000 0.07688 0.01151 0.04745 0.22270 -
 0.12000 0.11308 0.02458 0.06655 0.27097 -
 0.16000 0.14786 0.04148 0.09008 0.31084 -
 0.20000 0.18127 0.06155 0.11562 0.34528 -
 0.24000 0.21337 0.08420 0.14170 0.37579 -
 0.28000 0.24422 0.10891 0.16750 0.40330 -
 0.32000 0.27385 0.13524 0.19450 0.42839 -
 0.36000 0.30232 0.16279 0.22295 0.45149 -
 0.40000 0.32968 0.19121 0.25213 0.47291 -
 0.44000 0.35596 0.22021 0.28103 0.49288 -
 0.48000 0.38122 0.24953 0.30678 0.51158 -
 0.52000 0.40548 0.27895 0.33697 0.52916 -
 0.56000 0.42879 0.30829 0.36569 0.54574 -
 0.60000 0.45119 0.33737 0.39332 0.56142 -
 0.64000 0.47271 0.36607 0.41967 0.57629 -
 0.68000 0.49338 0.39428 0.44513 0.59041 -
 0.72000 0.51325 0.42190 0.47060 0.60386 -
 0.76000 0.53233 0.44885 0.49430 0.61667 -
 0.80000 0.55067 0.47507 0.51550 0.62891 -
 0.84000 0.56829 0.50052 0.53736 0.64060 -
 0.88000 0.58522 0.52516 0.55780 0.65180 -
 0.92000 0.60148 0.54896 0.57822 0.66253 -
 0.96000 0.61711 0.57191 0.59779 0.67281 -
 1.00000 0.63212 0.59399 0.61638 0.68269 -
 1.04000 0.64655 0.61521 0.63496 0.69218 -
 1.08000 0.66040 0.63557 0.65120 0.70130 -
 1.12000 0.67372 0.65507 0.66731 0.71008 -
 1.16000 0.68651 0.67373 0.68115 0.71853 -
 1.20000 0.69881 0.69156 0.69557 0.72668 -
 1.24000 0.71062 0.70857 0.70990 0.73453 -
 1.28000 0.72196 0.72480 0.72386 0.74210 -
 1.32000 0.73286 0.74025 0.73801 0.74941 -
 1.36000 0.74334 0.75495 0.74895 0.75646 -
 1.40000 0.75340 0.76892 0.75913 0.76328 -
 1.44000 0.76307 0.78220 0.77024 0.76986 -
 1.48000 0.77236 0.79480 0.78058 0.77623 -
 1.52000 0.78129 0.80675 0.79061 0.78238 -
 1.56000 0.78986 0.81807 0.79977 0.78833 -
 1.60000 0.79810 0.82880 0.80902 0.79410 -
 1.64000 0.80602 0.83895 0.81795 0.79967 -
 1.68000 0.81363 0.84855 0.82691 0.80508 -
 1.72000 0.82093 0.85763 0.83511 0.81031 -
 1.76000 0.82796 0.86621 0.84245 0.81538 -
 1.80000 0.83470 0.87431 0.85141 0.82029 -
 1.84000 0.84118 0.88196 0.85966 0.82505 -
 1.88000 0.84741 0.88917 0.86721 0.82967 -
 1.92000 0.85339 0.89597 0.87415 0.83414 -
 1.96000 0.85914 0.90238 0.87977 0.83849 -
 2.00000 0.86466 0.90842 0.88629 0.84270 -
$$
The culmulative intensity, N(Z), plot is diagnostic for both twinning and tNCS. For twinned data there are fewer weak reflections, therefore, N(Z) is sigmoidal for twinned data. However, if both twinning and tNCS are present, the effects may cancel each out. Therefore the results of the L-test and patterson test should be consulted
L test for twinning: (Padilla and Yeates Acta Cryst. D59 1124 (2003))
L statistic = 0.416 (untwinned 0.5 perfect twin 0.375)
Data has used to 67.93 - 3.50 A resolution
 Relation between L statistics and twinning fraction:
 Twinning fraction = 0.000 L statistics = 0.500:
 Twinning fraction = 0.100 L statistics = 0.440:
 Twinning fraction = 0.500 L statistics = 0.375:
The L test suggests data is twinned
All data regardless of I/sigma(I) has been included in the L test
$TABLE: L test for twinning:
$GRAPHS: cumulative distribution function for |L|, twin fraction of 0.18:0|1x0|1:1,2,3,4:
$$ |L| N(L) Untwinned Twinned $$
$$
0.0000 0.0000 0.0000 0.0000
0.0500 0.0666 0.0500 0.0749
0.1000 0.1339 0.1000 0.1495
0.1500 0.2007 0.1500 0.2233
0.2000 0.2648 0.2000 0.2960
0.2500 0.3263 0.2500 0.3672
0.3000 0.3870 0.3000 0.4365
0.3500 0.4475 0.3500 0.5036
0.4000 0.5070 0.4000 0.5680
0.4500 0.5646 0.4500 0.6294
0.5000 0.6206 0.5000 0.6875
0.5500 0.6756 0.5500 0.7418
0.6000 0.7274 0.6000 0.7920
0.6500 0.7768 0.6500 0.8377
0.7000 0.8229 0.7000 0.8785
0.7500 0.8643 0.7500 0.9141
0.8000 0.9025 0.8000 0.9440
0.8500 0.9355 0.8500 0.9679
0.9000 0.9628 0.9000 0.9855
0.9500 0.9842 0.9500 0.9963
1.0000 1.0000 1.0000 1.0000
$$
The Cumulative |L| plot for acentric data, where L = (I1-I2)/(I1+I2). This depends on the local difference in intensities. The difference operators used link to the neighbouring reflections taking into account possible tNCS operators.
Note that this estimate is not as reliable as obtained via the H-test or ML Britton test if twin laws are available. However, it is less prone to the effects of anisotropy than the H-test
Reference: Padilla, Yeates. A statistic for local intensity differences: robustness to anisotropy and pseudo-centering and utility for detecting twinning. Acta Cryst. D59, 1124-30, 2003.
Mean acentric moments I from input data:
 <I^2>/^2 = 1.818 (Expected = 2.000, Perfect Twin = 1.500)
 <I^3>/^3 = 5.843 (Expected value = 6.000, Perfect Twin = 3.000)
 <I^4>/^4 = 37.521 (Expected value = 24.000, Perfect Twin = 7.500)
$TABLE: Acentric Moments of I:
$GRAPHS: 2nd moment of I 1.818 (Expected value = 2, Perfect Twin = 1.5):0|0.112x0|5:1,2:
: 3rd & 4th Moments of I (Expected values = 6, 24, Perfect twin = 3, 7.5):0|0.112x0|36:1,3,4:
$$ 1/resol^2 <I**2> <I**3> <I**4> $$
$$
 0.006970 3.366 30.978 430.240
 0.015047 1.914 5.340 18.196
 0.021148 1.777 4.490 14.406
 0.026505 1.960 6.778 34.617
 0.031327 1.705 4.206 14.007
 0.035779 1.695 3.932 11.011
 0.039978 1.912 5.494 19.767
 0.043990 1.720 3.990 11.197
 0.047857 1.787 5.324 23.536
 0.051571 1.892 5.798 24.313
 0.055079 1.668 4.328 16.686
 0.058588 1.936 6.063 25.829
 0.061936 1.595 3.392 8.645
 0.065227 1.578 3.338 8.668
 0.068297 1.628 3.610 10.245
 0.071442 1.633 3.616 9.824
 0.074425 1.502 2.992 7.248
 0.077502 1.607 3.465 9.355
 0.080350 1.609 3.327 7.939
 0.083247 1.634 3.766 11.561
 0.086126 1.845 5.379 21.920
 0.088817 1.455 2.697 5.933
 0.091594 2.329 13.127 120.470
 0.094336 2.184 7.879 36.506
 0.096883 2.403 8.850 39.439
 0.099563 1.963 5.747 21.618
 0.102142 1.753 4.213 12.919
 0.104702 1.762 3.872 10.083
 0.107158 1.816 4.074 10.915
 0.109761 1.657 3.348 7.942
 0.112129 1.864 4.539 13.753
$$
First principles calculation has found 3 potential twinning operators
 # twinning operator score type
 0 k,h,-l 0.00 pm
 1 -h,-k,l 0.00 pm
 2 -h,h+k,-l 0.00 pm
 m merohedral
 pm pseudo-merohedral
The score gives an indication of the closure of the twinning operation. The lower the values
 the more higher the overlap.
The appearance of twinning operators only indicates that the crystal symmetry and lattice symmetry permit twinning. It does not mean that there is twinning present. Only the presence of statistics consistent with twinning gives a strong indicator.
Twinning operator based tests:
H-test: Cumulative plot of H=|I-T(I)|/(I-T(I)) for twin related reflections. This should be linear with slope 1/(1-2a).
$TABLE: H test for twinning
$GRAPHS: cumulative distribution function for |H| (operator k, h, -l) alpha = 0.42:0|1x0|1:1,2,3,4,5,6,7:
: cumulative distribution function for |H| (operator -h, -k, l) alpha = 0.39:0|1x0|1:1,2,3,4,5,6,8:
: cumulative distribution function for |H| (operator -h, h+k, -l) alpha = 0.39:0|1x0|1:1,2,3,4,5,6,9:
$$ |H| 0.4 0.3 0.2 0.1 0.0 k,h,-l -h,-k,l -h,h+k,-l$$
$$
0.00 0.0 0.0 0.0 0.0 0.0 0.00 0.00 0.00
0.05 - - - - - 0.68 0.43 0.45
0.10 - - - - - 0.88 0.72 0.72
0.15 - - - - - 0.94 0.86 0.86
0.20 - - - - - 0.96 0.92 0.93
0.25 - - - - - 0.98 0.95 0.96
0.30 - - - - - 0.98 0.97 0.97
0.35 - - - - - 0.99 0.98 0.98
0.40 - - - - - 0.99 0.98 0.98
0.45 - - - - - 0.99 0.99 0.99
0.50 - - - - - 1.00 0.99 0.99
0.55 - - - - - 1.00 0.99 0.99
0.60 - - - - - 1.00 0.99 0.99
0.65 - - - - - 1.00 1.00 1.00
0.70 - - - - - 1.00 1.00 1.00
0.75 - - - - - 1.00 1.00 1.00
0.80 - - - - - 1.00 1.00 1.00
0.85 - - - - - 1.00 1.00 1.00
0.90 - - - - - 1.00 1.00 1.00
0.95 - - - - - 1.00 1.00 1.00
1.00 5.0 2.5 1.67 1.25 1.0 1.00 1.00 1.00
$$
Britton plot: Plot of number of negative detwinned intensities.
$TABLE: Britton plot for twinning
$GRAPHS: aI1+(1-a)I2 > 0 (operator k, h, -l) alpha = 0.39:A:1,2:
: aI1+(1-a)I2 > 0 (operator -h, -k, l) alpha = 0.37:A:1,3:
: aI1+(1-a)I2 > 0 (operator -h, h+k, -l) alpha = 0.37:A:1,4:
$$ alpha k,h,-l -h,-k,l -h,h+k,-l$$
$$
0.00 0.00 0.00 0.00
0.03 0.00 0.00 0.00
0.05 0.00 0.00 0.00
0.07 0.00 0.00 0.00
0.10 0.00 0.00 0.00
0.12 0.00 0.00 0.00
0.15 0.00 0.00 0.00
0.17 0.00 0.00 0.00
0.20 0.00 0.00 0.00
0.23 0.00 0.00 0.00
0.25 0.00 0.00 0.00
0.28 0.00 0.00 0.00
0.30 0.00 0.01 0.01
0.33 0.00 0.01 0.01
0.35 0.01 0.01 0.01
0.38 0.01 0.01 0.01
0.40 0.01 0.02 0.02
0.42 0.02 0.04 0.04
0.45 0.03 0.07 0.07
0.47 0.08 0.14 0.14
$$
ML-Britton: Plot of number of negative detwinned intensities. The ML element corrects for the sigma in the observed intensity and for the effects of a single tNCS operator, if it is present.
$TABLE: ML-Britton test for twinning
$GRAPHS: aI1+(1-a)I2 > 0 (operator k, h, -l) alpha = 0.47:A:1,2:
: aI1+(1-a)I2 > 0 (operator -h, -k, l) alpha = 0.45:A:1,3:
: aI1+(1-a)I2 > 0 (operator -h, h+k, -l) alpha = 0.45:A:1,4:
$$ alpha k,h,-l -h,-k,l -h,h+k,-l$$
$$
0.00 0.26 0.26 0.26
0.03 -1922.56 -1919.46 -1918.94
0.05 -3987.00 -3977.78 -3974.50
0.07 -6157.25 -6136.57 -6131.95
0.10 -8447.11 -8411.39 -8406.03
0.12 -10877.36 -10824.58 -10818.17
0.15 -13468.83 -13394.48 -13385.80
0.17 -16243.41 -16140.40 -16129.30
0.20 -19226.63 -19084.18 -19071.17
0.23 -22450.29 -22252.97 -22240.06
0.25 -25953.54 -25681.19 -25672.06
0.28 -29785.11 -29408.61 -29408.35
0.30 -34009.00 -33478.99 -33493.97
0.33 -38707.08 -37931.58 -37972.35
0.35 -43982.57 -42782.44 -42863.82
0.38 -49953.87 -47966.38 -48105.22
0.40 -56711.13 -53196.56 -53411.44
0.42 -64141.85 -57712.76 -58018.04
0.45 -71453.32 -60141.97 -60541.82
0.47 -76835.02 -59398.04 -59888.16
$$
Twin fraction estimates based on global statistics:
 Twin fraction estimate from L-test: 0.18
 Twin fraction estimate from moments: 0.10
Twin fraction estimates by twinning operator
The following operator based twinning estimates analyse data with each of the possible twin operators. If twinning is present the most likely operator will have a low RTwin score (<I-T(I)>/<I+T(I)>) and estimates of the twin fraction above 0.
-------------------------------------------------------------------------------------------------
| operator | L-test | |Rtwin| | H-test | Britton | ML Britton |
-------------------------------------------------------------------------------------------------
| k, h, -l | Yes | 0.05 | 0.42 | 0.39 | 0.47 ( N/A ) |
| -h, -k, l | Yes | 0.08 | 0.39 | 0.37 | 0.45 ( N/A ) |
| -h, h+k, -l | Yes | 0.08 | 0.39 | 0.37 | 0.45 ( N/A ) |
-------------------------------------------------------------------------------------------------
TWINNING SUMMARY
Twinning fraction from H-test: 0.42
Twinning fraction from L-Test: 0.18
It is highly probable that your crystal is TWINNED.
 Please use twin refinement after your model is almost completed and R-free is below 40%
On Fri, Apr 14, 2017 at 12:10 PM, Eleanor Dodson <[log in to unmask]> wrote:
That twin factor list means the apparent crystal symmetry must be P6/mmm.

You say you only have 2 molecules in the asymmetric unit of P32,therefor there must only be one in SGs P32 21 P32 12

So I dont understand why you have PHASER results like this:
SOLU SET RFZ=4.4 TFZ=7.7 PAK=0 LLG=55 TFZ==9.6 LLG=350 TFZ==20.5 PAK=0 LLG=350 TFZ==20.5..

Why so many TFZ here - is that achieved after refinement or something? 

Eleanor

And what does the twinning analysis suggest?
On 14 April 2017 at 17:42, Keller, Jacob <[log in to unmask]> wrote:
As I mentioned off-list, it would be helpful to know how many types of search models you are searching with—how many different molecules are in the complex? It’s hard to interpret MR results otherwise.

Also, since the higher-symmetry SG works in MR, you should try to refine the model in that SG, with only two twin domains, refining twin fraction. I can guarantee that a good reviewer will have you do this (if not, then not a “good reviewer.”)

JPK

From: CCP4 bulletin board [mailto:[log in to unmask]] On Behalf Of Alex Lee
Sent: Friday, April 14, 2017 11:50 AM

To: [log in to unmask]
Subject: Re: [ccp4bb] Refmac5 twin refinement pushing Rfree surprisingly down
Thanks Eleanor, I tried MR for P32 21 and P32 12.
SG P3221: SOLU SET RFZ=5.3 TFZ=8.8 PAK=0 LLG=121 TFZ==11.2 LLG=944 TFZ==29.2 PAK=0 LLG=944 TFZ==29.2
   SOLU SPAC P 32 2 1
 
SG P3212:
Solution #1 annotation (history):
 
   SOLU SET  RFZ=4.4 TFZ=7.7 PAK=0 LLG=55 TFZ==9.6 LLG=350 TFZ==20.5 PAK=0 LLG=350 TFZ==20.5
 
   SOLU SPAC P 32 1 2
 
SG P32
SOLU SET RFZ=7.4 TFZ=10.4 PAK=0 LLG=187 TFZ==10.7 RF++ TFZ=17.0 PAK=0 LLG=436 TFZ==17.8 LLG=1715 TFZ==34.3 PAK=0
    LLG=1715 TFZ==34.3
   SOLU SPAC P 32
 
Based on TFZ and LLG, the P32 seems to be best. But I'll also try to refine and build P32 2 1 latter
On Fri, Apr 14, 2017 at 4:32 AM, Eleanor Dodson <[log in to unmask]> wrote:
First - four way twinning is possible but pretty rare for macromolecules

Pointless gives a very useful table of the CC agreement for each possible symmetry operator individually.

In this case with only two molecules in the asymmetric unit you you could only have a higher symmetry SG as

P32 21 P32 12 or P64

These would require as symmetry operators -

P32 21 - a three fold and a two fold k h -l

P32 12 - a three fold and a two fold -k -h -l

P64 - a six fold

If the scores for one set are better than the others you probably have that SG

However high degrees of twinning can disguise the symmetry scores of course..
On 14 April 2017 at 04:46, Keller, Jacob <[log in to unmask]> wrote:
Try MR with one copy in all space groups of PG 321/312 using Phaser. Going from PG 3 to PG 32 should halve the number of copies per ASU. You may have to re-process your data in the higher point group to do this.

Or you might actually have a tetartohedral twin, but just try with the higher-symmetry point group first, see what happens.

JPK

From: Alex Lee [mailto:[log in to unmask]m]
Sent: Thursday, April 13, 2017 11:32 PM

To: Keller, Jacob <[log in to unmask]>
Cc: [log in to unmask]
Subject: Re: [ccp4bb] Refmac5 twin refinement pushing Rfree surprisingly down
Hi Keller,

Thanks for the suggestions! I only have two copies in ASU at SG P32. Zanuda also suggests P32 is the best SG.
On Thu, Apr 13, 2017 at 8:12 PM, Keller, Jacob <[log in to unmask]> wrote:
Yes, this was my case exactly—it looks like there are two pairs of coupled twin domains: a,c and b,d. Assuming you have multiple copies of your model in the same ASU, try doing MR in higher symmetry space groups of point group 312 or 321, like P3212 etc. There is this handy page with all the space groups and their possible twin operators: http://www.ccp4.ac.uk/html/twinning.html.

The twin fractions indicate a high twin fraction—~46% if actually hemihedral!

Also take a look at the paper I referenced for more info. I can send you a .pdf if you need me to.

Please let me know how it works out—I am interested in these types of things!

JPK

From: Alex Lee [mailto:[log in to unmask]m]
Sent: Thursday, April 13, 2017 9:08 PM
To: Keller, Jacob <[log in to unmask]>
Cc: [log in to unmask]

Subject: Re: [ccp4bb] Refmac5 twin refinement pushing Rfree surprisingly down
Hi Keller,

I do not how to check twin fraction after Refmac (I guess it's somewhere in log file). From the log file it seems I have four twin domain:
Twin operators with estimated twin fractions ****
 
Twin operator:  H,  K,  L: Fraction = 0.275; Equivalent operators:  K, -H-K,  L; -H-K,  H,  L
Twin operator: -K, -H, -L: Fraction = 0.228; Equivalent operators: -H,  H+K, -L;  H+K, -K, -L
Twin operator:  K,  H, -L: Fraction = 0.270; Equivalent operators:  H, -H-K, -L; -H-K,  K, -L
Twin operator: -H, -K,  L: Fraction = 0.228; Equivalent operators: -K,  H+K,  L;  H+K, -H,  L
On Thu, Apr 13, 2017 at 4:36 PM, Keller, Jacob <[log in to unmask]> wrote:

What was the refined twin fraction after Refmac? It’s much more accurate than initial tests. Also, how many twin domains do you have? If you have many, it might be a higher space group but with less twinning. I recently had a case in which apparent tetartohedral (four-domain) twinning in P32 was really hemihedral (two-domain) twinning in P3212:

Acta Cryst. (2017). D73, 22-31
https://doi.org/10.1107/S2059798316019318

Jacob

From: CCP4 bulletin board [mailto:[log in to unmask]" target="_blank">[log in to unmask]] On Behalf Of Eleanor Dodson
Sent: Thursday, April 13, 2017 3:11 PM
To: [log in to unmask]
Subject: Re: [ccp4bb] Refmac5 twin refinement pushing Rfree surprisingly down

Twin refinement cannot be compared directly to untwinned - the R factors are between different parameters - without twinning it is assumed you have an amplitude obtained more or less from sqrt(I But for a twinned data set that I is actually [ I1 + twin_factor I2 ] so the amplitude is not really correct and twinned refinement will give a much better estimate.

However you need to be careful that you have assigned the same FreeR flag to reflection pair related by the twin law. The modern program in the CCP4 data reduction pipeline looks after this pretty automatically - all possible symmetry equivalents are assigned the same FreeR but older software did not do this..

You can check it by looking at some twin equivalents - in SG P32 these could be h k l and -h, -k, l or h k l and k h -l or h k l and -k, -h, -l .

Ideally they all should have the same Free R flag..

Eleanor

PS - the acid test is: Do the maps look better?

E

On 13 April 2017 at 19:52, Robbie Joosten <[log in to unmask]> wrote:

Hi Alex,

You are not giving the number after refinement without the twin refinement. Nevertheless, R-free drops like this are not unheard of. You should check your Refmac log file, it would warn you of potential space group errors. Refmac will also give you a refined estimate of the twin fraction.

Cheers,

Robbie

Sent from my Windows 10 phone

Van: Alex Lee
Verzonden: donderdag 13 april 2017 19:19
Aan: [log in to unmask]
Onderwerp: [ccp4bb] Refmac5 twin refinement pushing Rfree surprisingly down

Dear All,

I have a protein/dna complex crystal and data collected at 3A and another set at 2.8A, space group P32. L test shows twinning (fraction around 0.11). The structure solved by MR and model building of the complex finish (no solvent built yet, I do not think it's good to build solvent in such low resolution data).

I did Refmac5 to refine my structure (restraint refinement) with or without twinning, to my surprise, the Rfree drops a lot after twin refinement of two data sets. Summary below:

2.8A dataset: before twin refine 34%, 29%; after twin refine:24%, 19%

3A dataset: before twin refine 30%;26%; after refine 25%, 18%

I know that a lot of threads in CCP4bb talking about Rfree after twin refine and Rfree without twin refine can not compare directly. By drop R free this much by twin refine, it gives me a feeling of too good to be true (at such low resolution with such good Rfree, maybe overrefined a lot?), but from the density map after twin refine, it does seem better than no twin refine map.

I do not know if reviewers are going to challenge this part.

Any input is appreciated.